Vector Addition and Subtraction

CSEC Mathematics: Combining Vectors

Essential Understanding: Vectors can be added or subtracted to find resultant vectors. This is essential for solving problems in navigation, physics, and geometry. Understanding the Triangle Law and Parallelogram Law will help you visualize and calculate vector operations.

Key Skill: Adding Column Vectors
Key Skill: Subtracting Vectors
Exam Focus: Triangle & Parallelogram Laws

Adding Vectors Algebraically

When adding vectors in column form, simply add the corresponding components:

Vector Addition Formula

\[\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}\]

Add x-components together, add y-components together

Worked Example 1: Basic Addition

Calculate: \(\begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \end{pmatrix}\)

1
Add x-components: \(3 + 2 = 5\)
2
Add y-components: \(5 + (-1) = 4\)
3
Write the result: \(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\)

Answer: \(\begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}\)

Subtracting Vectors Algebraically

When subtracting vectors, subtract the corresponding components:

Vector Subtraction Formula

\[\begin{pmatrix} a \\ b \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a - c \\ b - d \end{pmatrix}\]

Subtract x-components, subtract y-components

Worked Example 2: Vector Subtraction

Calculate: \(\begin{pmatrix} 7 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix}\)

1
Subtract x-components: \(7 - 4 = 3\)
2
Subtract y-components: \(2 - 6 = -4\)
3
Write the result: \(\begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

Answer: \(\begin{pmatrix} 7 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

Remember: Subtraction as Addition

\(\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})\)

Subtracting a vector is the same as adding its negative (inverse) vector.

Scalar Multiplication of Vectors

Multiplying a vector by a scalar (number) multiplies each component:

Scalar Multiplication

\[k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}\]

Worked Example 3: Scalar Multiplication

Calculate: \(3\begin{pmatrix} 2 \\ -4 \end{pmatrix}\)

Solution: \(3\begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \times 2 \\ 3 \times (-4) \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \end{pmatrix}\)

Geometric Interpretation: Triangle Law

The Triangle Law of Vector Addition

If two vectors are represented by two sides of a triangle taken in order, then their resultant (sum) is represented by the third side taken in the opposite direction.

a b a + b A B C

To add a and b: Place the tail of b at the head of a. The resultant goes from the tail of a to the head of b.

Geometric Interpretation: Parallelogram Law

The Parallelogram Law of Vector Addition

If two vectors are represented by two adjacent sides of a parallelogram, then their resultant is represented by the diagonal passing through their common point.

a b a + b O

Both vectors start from the same point. The diagonal of the parallelogram gives the resultant.

Interactive Vector Addition Lab

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Vector Addition Calculator

Enter two vectors and see their addition visualized using the Triangle Law!

Vector a

\(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\)

Vector b

\(\begin{pmatrix} 2 \\ 4 \end{pmatrix}\)

a + b (Resultant)

\(\begin{pmatrix} 5 \\ 6 \end{pmatrix}\)

Vector Subtraction Geometrically

Understanding a - b

To subtract vector b from a, we add the negative of b:

\(\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})\)

a -b a - b

Position Vectors

A position vector describes the position of a point relative to the origin. If point P has coordinates (x, y), its position vector is:

\[\overrightarrow{OP} = \begin{pmatrix} x \\ y \end{pmatrix}\]

Worked Example 4: Finding a Vector Between Two Points

Given: Point A(2, 3) and Point B(7, 1)

Find: \(\overrightarrow{AB}\)

1
Formula: \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)
2
Position vectors: \(\overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\), \(\overrightarrow{OB} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}\)
3
\(\overrightarrow{AB} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 7-2 \\ 1-3 \end{pmatrix} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\)

Answer: \(\overrightarrow{AB} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\)

Important Formula

To find the vector from point A to point B:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)

Or simply: subtract coordinates of A from coordinates of B

CSEC Practice Questions

Test Your Understanding

1
Calculate: \(\begin{pmatrix} 4 \\ -3 \end{pmatrix} + \begin{pmatrix} -2 \\ 7 \end{pmatrix}\)
\(\begin{pmatrix} 2 \\ 4 \end{pmatrix}\)
\(\begin{pmatrix} 6 \\ 4 \end{pmatrix}\)
\(\begin{pmatrix} 2 \\ -10 \end{pmatrix}\)
\(\begin{pmatrix} -6 \\ 10 \end{pmatrix}\)
Solution: \(\begin{pmatrix} 4 + (-2) \\ -3 + 7 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\)
2
Calculate: \(\begin{pmatrix} 5 \\ 8 \end{pmatrix} - \begin{pmatrix} 3 \\ 11 \end{pmatrix}\)
\(\begin{pmatrix} 8 \\ 19 \end{pmatrix}\)
\(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\)
\(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\)
Solution: \(\begin{pmatrix} 5 - 3 \\ 8 - 11 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\)
3
If \(\mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}\), find \(2\mathbf{a} + \mathbf{b}\)
\(\begin{pmatrix} 2 \\ 16 \end{pmatrix}\)
\(\begin{pmatrix} 5 \\ 11 \end{pmatrix}\)
\(\begin{pmatrix} 3 \\ 13 \end{pmatrix}\)
\(\begin{pmatrix} 1 \\ 13 \end{pmatrix}\)
Solution:
\(2\mathbf{a} = 2\begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 10 \end{pmatrix}\)
\(2\mathbf{a} + \mathbf{b} = \begin{pmatrix} 4 \\ 10 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 13 \end{pmatrix}\)
4
Point P has coordinates (3, 4) and point Q has coordinates (8, -2). Find \(\overrightarrow{PQ}\)
\(\begin{pmatrix} 11 \\ 2 \end{pmatrix}\)
\(\begin{pmatrix} -5 \\ 6 \end{pmatrix}\)
\(\begin{pmatrix} 5 \\ -6 \end{pmatrix}\)
\(\begin{pmatrix} -5 \\ -6 \end{pmatrix}\)
Solution: \(\overrightarrow{PQ} = \begin{pmatrix} 8 - 3 \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -6 \end{pmatrix}\)
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CSEC Examination Tips

  • Be careful with signs! Double-check when subtracting negative numbers.
  • Draw diagrams when dealing with position vectors - it helps visualize the problem.
  • For \(\overrightarrow{AB}\): Remember it's always "End minus Start" (B minus A).
  • Common CSEC question: Given position vectors, find the vector between two points and its magnitude.
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